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jbritto
 2 years ago
What is the limit of (1+cosx)/sinx when x approaches 0
jbritto
 2 years ago
What is the limit of (1+cosx)/sinx when x approaches 0

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zugzwang
 2 years ago
Best ResponseYou've already chosen the best response.0l'Hôpital's rule, are you allowed to use that?

zugzwang
 2 years ago
Best ResponseYou've already chosen the best response.0Wait, I panicked... no l'Hôpital's rule, I'm sorry

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2How can you use lhospital here?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 0}\frac{1+\cos(x)}{\sin(x)}\] Recall the following: \[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \text{ , } \lim_{x \rightarrow 0}\frac{\cos(x)1}{x}=0\] Let's see if we can use that here: \[\lim_{x \rightarrow 0}\frac{\cos(x)+1}{\sin(x)} \cdot \frac{\cos(x)1}{\cos(x)1}\] \[\lim_{x \rightarrow 0} \frac{\cos^2(x)1}{\sin(x)(\cos(x)1)}\] \[\lim_{x \rightarrow 0}=\frac{\sin^2(x)}{\sin(x)(\cos(x)1)}\] \[\lim_{x \rightarrow 0}\frac{\sin(x)}{\cos(x)1}\] Now try it :)

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2Using L'hospital you will get something easier to look at :)

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.2We didn't actually need any of those things I told you to recall above :)

jbritto
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks. It still becomes 1/0 which is undefined ??

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0its not 1/0 but its 0/0 @jbritto sin0 = 0 cos0 =1

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0anyways,,there is an alternative also.. 1+ cosx = 1 + 2 cos^2 (x/2) 1 = 2cos^2 (x/2) and sinx = 2 sin(x/2) cos(x/2) so whole expression is 2 cos^2(x/2) / 2 sin(x/2) cos(x/2) = cos(x/2)/ sin(x/2) = cot (x/2) not plug in x=0

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0oh,,so you were saying it becomes 1/0 !! ohh,,am sorry.. in that case either the limit is undefined or you maybe made a typing error//

jbritto
 2 years ago
Best ResponseYou've already chosen the best response.0yes. applying LHosp rule, it will become cosx/sinx which is 1/0 undefined
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